19 May
2003
19 May
'03
11:02 a.m.
Dnia wto 13. maja 2003 09:58, Simon Peyton-Jones napisał:
Think of it like this. Should this be acceptable?
f :: a -> a -> a f x y = x && y
No, because the type (forall a. a->a->a) is plainly more general than the actual function.
It's not the same. The type doesn't constrain a by any class, so it would look like any type fits, which is not true.
bar :: (Foo Char t) => t
In this case the type seems OK: for any type t, *if* Foo Char t, then bar can be used as type t. -- __("< Marcin Kowalczyk \__/ qrczak@knm.org.pl ^^ http://qrnik.knm.org.pl/~qrczak/